# NCERT Solutions for Class 11 Chemistry

Find 100% accurate solutions for NCERT Class XI Chemistry. All questions have been explained and solved step-by-step, to help you understand thoroughly. Free Download option available!

1.    (a) NaH2PO

Let the oxidation number of P be x.

We know that,

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = -2

Then, we have

1(+1) + 2(+1) +1 (×) + 4(-2) = 0

⇒ 1 + 2 + × – 8 = 0

⇒ x = +5

Hence, the oxidation number of P is +5.

(b) NaHSO4

Then, we have: 1(1+1) +1 (+1) +1(x) +4(-2) = 0

= 1+1+ x – 8 = 0

= x = +6

Hence, the oxidation number of S is + 6.

(c) H4P2O7

Then, we have: 4(+1) +2(x) +7(-2)

= 4 + 2x – 14 = 0

= 2x = +10 x = +5

Hence, the oxidation number of P is + 5.

(d) K2MnO4

Then, we have: 2(+1) +x +4(-2) = 0

= 2 + x – 8 = 0

= x = +6

Hence, the oxidation number of Mn is + 6.

(e) CaO2

Then, we have: (+2) +2(x) = 0

= 2 + 2x = 0

= x = -1

Hence, the oxidation number of O is -1.

(f) NaBH4

Then, we have

1(+1)+1(x)+4(-1) = 0

⇒ 1 + x – 4 = 0

⇒ x = +3

Hence, the oxidation number of B is +3.

(g) H2S2O7

Then, we have

2(+1)+2(x)+7(-2) = 0

⇒ 2 + 2x – 14 = 0

⇒ 2x = 12

⇒ x = +6

Hence, the oxidation number of S is +6.

(h) KAI (SO4)2.12 H2O

Then, we have

1(+1)+1(+3)+2(x)+8(-2)+24(+1)+12(-2) = 0

⇒ 1 + 3 + 2x – 16 + 24 = 0

⇒ 2x = 12

⇒ x = +6

Or,

We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the water molecule, we have

1(+1)+1(+3)+2(x)+8(-2) = 0

⇒ 1 + 3 + 2x – 16 = 0

⇒ 2x = 12

⇒ x = +6

Hence, the oxidation number of S is +6.

2.     (a) KI3

In KI3, the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is . However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI3 to find the oxidation states.

In a KI3 molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

Hence, in a KI3 molecule, the O.N. of the two I atoms forming the I2 molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is -1.

(b) H2S4O6

However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.

The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.

(c) Fe3O4

On taking the O.N. of O as -2, the O.N. of Fe is found to be . However, O.N. cannot be fractional.

Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of +3.

(d) CH3CH2OH

2 (x) + 6 (+1) + 1 (-2) = 0

or, 2x + 4 = 0

or, x = -2

Hence, the O.N. of C is -2.

(e) CH3COOH

2 (x) + 4 (+1) + 2 (-2) = 0

or, 2x = 0

or, x = 0

However, 0 is average O.N. of C. The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits the oxidation states of +2 and -2 in CH3COOH.

3.    (a) CuO(s) + H2(g) → Cu(s) + H2O(g)

Let us write the oxidation number of each element involved in the given reaction as:

Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu. Also, the oxidation number of H increases from 0 in  to +1 in  i.e.,  is H2 to  +1 in H2O i.e., His oxidized to H2O . Hence, this reaction is a redox reaction.

(b) Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)

Let us write the oxidation number of each element in the given reaction as:

Here, the oxidation number of Fe decreases from +3 in Fe2O3 to 0 in Fe i.e., Fe2O3 is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in COi.e., CO is oxidized to CO2. Hence, the given reaction is a redox reaction.

(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3AlCl3(s)

The oxidation number of each element in the given reaction can be represented as:

In this reaction, the oxidation number of B decreases from +3 in BCl3 to -3 in B2H6. i.e., BCl3 is reduced to B2H6. Also, the oxidation number of H increases from -1 in LiAlH4 to +1 in B2H6 i.e., LiAlH4 is oxidized to B2H6. Hence, the given reaction is a redox reaction.

(d) 2K(s) + F2(g) →2K + F-(s)

The oxidation number of each element in the given reaction can be represented as:

In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in F2 to -1 in KF i.e., F2 is reduced to KF.

Hence, the above reaction is a redox reaction.

(e) 4NH3(g) + 5O2(g) → 4NO(s) + 6H2O(g)

The oxidation number of each element in the given reaction can be represented as:

Here, the oxidation number of N increases from -3 in NH3 to +2 in NO. On the other hand, the oxidation number of O2 decreases from 0 in Oto -2 in NO and H2O i.e., O2 is reduced. Hence, the given reaction is a redox reaction.

4.    Let us write the oxidation number of each atom involved in the given reaction above its symbol as:

Here, we have observed that the oxidation number of F increases from 0 in F2 to +1 in HOF. Also, the oxidation number decreases from 0 in F2  to -1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.

5.

However, the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S cannot be more than +6.

The structure of H2SO4 is shown as follows:

Now, 2(+1)+1(x) + 3(-2) + 2(-1) = 0

⇒ 2 + x – 6 – 2 = 0

⇒ x = +6

Therefore, the O.N of S is +6.

(ii)

2(x) + 7(-2) = -2

⇒ 2x – 14 = -2

⇒ x = +6

Here, there is no fallacy about the O.N. of Cr in Cr2O2-7

The structure of Cr2O2-is shown as follows:

Here, each of the two Cr atoms exhibits the O.N. of +6.

(iii)

1(x) +3(-2) = -1

⇒ x – 6 = -1

⇒ x = +5

Here, there is no fallacy about the O.N. of N in NO3

The structure of NO3 is shown as follows:

The N atom exhibits the O.N. of +5.

6.    (a) Mercury (II) chloride: HgCl2

(b) Nickel (II) sulphate: NiSO4

(c) Tin (IV) oxide: SnO2

(d) Thallium (I) sulphate: Tl2SO4

(e) Iron (III) sulphate: Fe2 (SO4)3

(f) Chromium (III) oxide: Cr2O3

7.     The substances were carbon can exhibit oxidation states from -4 to +4 are listed in the following table.

The substances where nitrogen can exhibit oxidation states from -3 to +5 are listed in the following  table.

8.   In sulphur dioxide (SO2), the oxidation number (O.N.) of S is +4 and the range of the O.N. that S can have is from +6 to -2.

Therefore, SO2 can act as an oxidising as well as a reducing agent.

In hydrogen peroxide (H2O2), the O.N. of O is -1 and the range of the O.N. that O can have is from 0 to -2. O can sometimes also attain the oxidation numbers +1 and +2. Hence, H2O2 can act as an oxidising as well as a reducing agent.

In ozone (O3), the O.N. of O is zero and the range of the O.N. that O can have is from 0 to -2. Therefore, the O.N. of O can only decrease in this case. Hence, O3 acts only as an oxidant.

In nitric acid (HNO3), the O.N. of N is +5 and the range of the O.N. that N can have is from +5 to -3. Therefore, the O.N. of N can only decrease in this case. Hence, HNO3 acts only as an  oxidant.

9.    (a) The process of photosynthesis involves two steps.

Step 1:

H2O decomposes to give H2 and O2.

2H2O(l) → 2H2(g) + O2(g)

Step 2:

The H2 produced in step 1 reduces CO2, thereby producing gluocse (C6H12O6) and H2O.

6CO2(g) + 12H2(g) → C6H12O6(s) + 6H2O(l)

Now, the net reaction of the process is given as:

It is more appropriate to write the reaction as given above because water molecules are also produced in the process of photosynthesis.

The path of this reaction can be investigated by using radioactive H2O in place of H2O.

(b)  is produced from each of the two reactants O3 and H2O. For this reason, O2 is written  twice.

The given reaction involves two steps. First, O3 decomposes to form O2 and O. In the second step, H2O reacts with the O produced in the first step, thereby producing H2O  and O2.

The path of this reaction can be investigated by using H2O218  or O218.

10.   The oxidation state of Ag in AgF2 is +2. But, +2 is an unstable oxidation state of Ag. Therefore,  whenever AgF2 is formed, silver readily accepts an electron to form Ag+. This helps to bring the oxidation state of Ag down from +2 to a more stable state of +1. As a result, AgF2  acts as a very strong oxidizing agent.

11.   Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. This can be illustrated as follows:

(i) P4 and F2 are reducing and oxidising agents respectively.

If an excess of P4 is treated with F2, then PF3 will be produced, wherein the oxidation number (O.N.) of P is +3.

However, if P4 is treated with an excess of F2, then PF5 will be produced, wherein the O.N. of P is +5.

(ii) K acts as a reducing agent, whereas O2 is an oxidising agent.

If an excess of K reacts with O2, then K2O will be formed, wherein the O.N. of O is -2.

However, if K reacts with an excess of O2, then K2O2 will be formed, wherein the O.N. of O is -1.

(iii) C is a reducing agent, while O2 acts as an oxidising agent.

If an excess of C is burnt in the presence of insufficient amount of O2, then CO will be produced, wherein the O.N. of C is +2.

On the other hand, if C is burnt in an excess of O2, then CO2 will be produced, wherein the O.N. of C is +4.

12.   (a) In the manufacture of benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant because of the following reasons.

(i) In a neutral medium, OH ions are produced in the reaction itself. As a result, the cost of adding an acid or a base can be reduced.

(ii) KMnO4 and alcohol are homogeneous to each other since both are polar. Toluene and alcohol are also homogeneous to each other because both are organic compounds. Reactions can proceed at a faster rate in a homogeneous medium than in a heterogeneous medium.

Hence, in alcohol, KMnO4 and toluene can react at a faster rate.

The balanced redox equation for the reaction in a neutral medium is give as below:

(b) When conc. H2SO4 is added to an inorganic mixture containing bromide, initially HBr is produced. HBr, being a strong reducing agent reduces H2SO4 to  SO2 with the evolution of red vapour of bromine.

But, when conc. H2SO4 is added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, being a weak reducing agent, cannot reduce H2SO4 to SO2.

2NaCl + 2H2SO4 → 2NaHSO+ 2HCl

13.   (a) Oxidised substance- C6H6O2

Reduced substance- AgBr

AgBr Oxidising  agent – AgBr

Reducing agent – C6H6O2

(b) Oxidised substance – HCHO

Reduced substance -[Ag(NH3)2]+

Oxidising agent -[Ag(NH3)2]+

Reducing agent-HCHO

(c) Oxidised substance-HCHO

Reduced substance-Cu2+

Oxidising agent – Cu2+

Reducing agent -HCHO

(d) Oxidised substance – N2H4

Reduced substance- H2O2

Oxidising  agent- H2O2

Reducing agent-N2H4

(e) Oxidised substance

Reduced substance – PbO2

Oxidising agent – PbO2

Reducing agent –Pb

14.   The average oxidation number (O.N.) of S in S2O32- is +2. Being a stronger oxidising agent than I2, Broxidises S2O32- to SO42-, in which the O.N. of S is +6. However, I2is a weak oxidising agent. Therefore, it oxidises S2O32- to S4O62-, in which the average O.N. of S is only +2.5. As a result, S2O32- reacts differently with iodine and bromine.

15.    F2 can oxidize Cl to Cl2, Br to Br2, and I to I2 as:

F2(aq) + 2Cl(s) → 2F(aq) + Cl2(g)

F2(aq) + 2Br(aq) → 2F(aq) + Br(g)

F2(aq) + 2l(s) → 2F(aq) + I2(g)

On the other hand, Cl2, Brand I2 cannot oxidize F to F2. The oxidizing power of halogens increases in the order of I2 < Br2 < Cl2 < F2. Hence, fluorine is the best oxidant among halogens.

HI and HBr can reduce H2SO4 to SO2, but HCl and HF cannot. Therefore, Hi and HBr are stronger reductants than HCl and HF.

2HI + H2SO4 → I2 + SO2 + 2H2O

2HBr + H2SO4 → Br2 + SO2 + 2H2O

Again, I can reduce Cu2+ to Cu+, but Br cannot.

4I(aq) + 2Cu2+(aq) → Cu2I2(s) + I2(aq)

Hence, hydroiodic acid is the best reductant among hydrohalic compounds.

Thus, the reducing power of hydrohalic acids increases in the order of HF < HCl < HBr < HI.

16.   The given reaction occurs because XeO64- oxidises being an oxidising agent and F reduces being an reducing agent to the XeO64-.

In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in XeO64- to +6 in XeO3 and the O.N. of F increases from -1 in F to O in F2.

Hence, we can conclude that Na4XeO6 is a stronger oxidising agent than F.

17.   Ag+ and Cu2+ act as oxidising agents in reactions (a) and (b) respectively.

In reaction (c), Ag+ oxidises C6H5CHO to C6H5COO, but in reaction (d), Cu2+ cannot oxidise C6H5CHO.

Hence, we can say that Ag+ is a stronger oxidising agent than Cu2+.

18.   (a) Step 1: The two half reactions involved in the given reaction are:

Oxidation half reaction:

Reduction half reaction:

Step 2:

Balancing I in the oxidation half reaction, we have:

2I(aq) → I2(g)

Now, to balance the charge, we add 2e to the RHS of the reaction.

2I(aq) → I2(g) + 2e

Step 3:

In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3 electrons are added to the LHS of the reaction.

MnO4 (aq) + 3e → MnO2(aq)

Now, to balance the charge, we add 4 OH ions to the RHS of the reaction as the reaction is taking place in a basic medium.

MnO4 (aq) + 3e → MnO2(aq) + 4OH

Step 4:

MnO4 (aq) +2H2O + 3e → MnO2(aq) + 4OH

In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS.

Step 5:

Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:

6I(aq) → 3I2(s) + 6e

2MnO4(aq) + 4H2O + 6e → 2MnO2(g) + 8OH(aq)

Step 6:

Adding the two half reactions, we have the net balanced redox reaction as:

6I(aq) + 2MnO4(aq) + 4H2O → 3I2(s) + 2MnO2(s) + 8OH(aq)

(b) Following the steps as in part (a), we have the oxidation half reaction as:

SO2(aq) + 2H2O(l) → HSO4(aq) + 3H+(aq) + 2e(aq)

And the reduction half reaction as:

MnO4(aq) + 8HO+(aq) + 5e → Mn2+(aq) + 4H2O(l)

Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as:

2MnO4(aq) + 5SO2(aq) + 2H2O(l) + H+(aq) → 2Mn2+(g) + 5HSO4(aq)

Following the steps as in part (a), we have the oxidation half reaction as:

Fe2+(aq) → Fe3+(aq) + e

And the reduction half reaction as:

H2O2 (aq) + 2H(aq) + 2e → 2H2O(l)

Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

H2O2 (aq) + 2Fe2+(aq) + 2H+ → 2Fe3+(aq) + 2H2O(l)

(d) Following the steps as in part (a), we have the oxidation half reaction as:

SO2(g) + 2H2O(l) → SO2-4(aq) + 4H+(aq) + 2e

And the reduction half reaction as:

CrO2-7(aq) + 14H+(aq) + 6e → 2Cr3+(aq) + 7H2O(l)

Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

CrO2-7(aq) + 14H+(aq) + 3SO2(g) + 2H+(aq) → 2Cr3+(aq) + 3SO2-4(aq)

19.   (a) The O.N. (oxidation number) of P decreases from 0 in P4 to O.N. 3 in PH3 and increases from 0 in P4 to + 2 in HPO1. Hence, P4 acts both as an oxidizing agent and a reducing agent in this reaction.

Ion electron method:

The oxidation half equation is:

P4(s) → HPO2(aq)

The P atom is balanced as:

The O.N. is balanced by adding 8 electrons as:

P4(s) → 4HPO2(aq) + 8e

The charge is balanced by adding 4H2O as:

P4(s) + 12OH(aq) → 4HPO2(aq) + 4H2O(l) + 8e……..(i)

The O and H atoms are already balanced.

The reduction half equation is:

P4(s) → PH3(g)

The O.N. is balanced by adding 12 electrons as:

P4(s) → 12e + 4PH3(g)

The charge is balanced by adding 12OH as:

P4(s) → 12e → 4PH3(g) + 12OH(aq)

The O and H atoms are balanced by adding as:

P4(s) → 12e → 4PH3(g) + 12OH(aq)      (ii)

By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:

5P4(s) + 12H2O(l) + 12OH(aq) → 8PH3(g) + 12HPO2(aq)

(b)

The oxidation number of N increases from -2 in N2H4 to +2 NO and the oxidation number of Cl decreases from + 5 in ClO3 to -1 in Cl. Hence, in this reaction, N2H4 is the reducing agent and ClOis the oxidizing agent.

Ion-electron method:

The oxidation half equation is:

The N atoms are balanced as:

N2H4(l) → 2NO(l) + 8e

The oxidation number is balanced by adding 8 electrons as:

N2H4(l) → 2NO(g) + 8e

The charge is balanced by adding 8 OH ions as:

N2H4(l)+ 8OH(aq) → 2NO(g) + 8e

The O atoms are balanced by adding 6H2O as:

N2H4(l) + 8OH(aq) → 2NO(g) + 6H2O(l) + 8e …….(i)

The reduction half equation is:

The oxidation number is balanced by adding 6 electrons as:

ClO3(aq) + 6e → Cl(aq)

The charge is balanced by adding 6O H ions as:

ClO3(aq) + 6e → Cl(aq) + 6OH(aq)

The O atoms are balanced by adding 3H2O as:

ClO3(aq) + 3H2O(l) + 6e → Cl(aq) + 6OH(aq) ……(ii)

The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as:

3N2H4(l) + 4ClO3(aq) → 6NO(g) + 4Cl(aq) + 6H2O(l)

Oxidation number method:

Total decrease in oxidation number of N = 2 × 4 = 8

Total increase in oxidation number of Cl = 1 × 6 = 6

On multiplying N2H4 with 3 and ClO3 with 4 to balance the increase and decrease in O.N., we get:

3N2H4(l) + 4ClO3(aq) → NO(g) + Cl(aq)

The N and Cl atoms are balanced as:

3N2H4(l) + 4ClO3(aq) → 6NO(g) + 4Cl(aq)

The O atoms are balanced by adding 6H2O as:

3N2H4(l) + 4ClO3(aq) → 6NO(g) + 4Cl(aq) + 6H2O(l)

This is the required balanced equation.

(c)

The oxidation number of Cl decreases from + 7 in Cl2O7 to + 3 in ClOand the oxidation number of O increases from -1 in H2O2 to zero in O2. Hence, in this reaction,  Cl2Ois the oxidizing agent and H2O2 is the reducing agent.

In-electron method:

The oxidation half equation is:

The oxidation number is balanced by adding 2 electrons as:

H2O2(aq) → O2(aq) + 2e

The charge is balanced by adding 2OH ions as:

H2O2(aq) + 2OH(aq) → O2(g) + 2e

The oxygen atoms are balanced by adding 2H2O as:

H2O2(aq) + 2OH(aq) → O2(g) + 2H2O(l) + 2e  …..(i)

The reduction half equation is:

The Cl atoms are balanced as:

Cl2O7(g) → 2ClO2(aq)

The oxidation number is balanced by adding 8 electrons as:

The charge is balanced by adding 6O H as:

Cl2O7(g) + 8e → 2ClO2(aq) + 6OH(aq)

The oxygen atoms are balanced by adding 3H2O as:

Cl2O7 + 3H2O(l) + 8e → 2ClO2(aq) + 6OH(aq)  ……(ii)

The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as: Cl2O7(g) + 4H2O(l) + 20H(aq) → 2ClO2(aq) + 6OH(aq) + 5H2O(l)

Oxidation number method:

Total decrease in oxidation number of Cl2O7 = 4 × 2 = 8

Total increase in oxidation number of H2O2 = 2 × 1 = 2

By multiplying H2O2 and Owith 4 to balance the increase and decrease in the oxidation number, we get:

Cl2O7(g) + 4H2O2(aq) → ClO2(aq) + 4O2(g)

The Cl atoms are balanced as:

Cl2O7(g) + 4H2O2(aq) → 2ClO2(aq) + 4O2(g)

The O atoms are balanced by adding 3H2O as:

Cl2O7(g) + 4H2O2(aq) → 2ClO2(aq) + 4O2(g) + 3H2O(l)

The H atoms are balanced by adding 2 OH and 2H2O( as:

Cl2O7(g) + 4H2O2(aq) + 2OH(aq) → 2ClO2(aq) + 4O2(g) + 5H2O(l)

This is the required balanced equation.

20.   The oxidation numbers of carbon in (CN)2, CN and CNO are +3, +2 and +4 respectively. These are obtained as shown below:

Let the oxidation number of C be x.

(CN)2

2(x -3) = 0

∴ x = 3

CN

x – 3 = -1

∴ x = 2

CNO

x – 3 – 2 = -1

∴ x = 4

The oxidation number of carbon in the various species is:

It can be easily observed that the same compound is being reduced and oxidised simultaneously in the given equation. Reactions in which the same compound is reduced and oxidised are known as disproportionation reactions. Thus, it can be said that the alkaline decomposition of cyanogen is an example of disproportionation reaction.

21.   The given reaction can be represented as:

Mn3+(aq) → Mn2+(aq) + MnO2(s) + H+(aq)

The oxidation half equation is:

The oxidation number is balanced by adding one electron as:

Mn3+(aq) → MnO2(s) + e(aq)

The charge is balanced by adding 4 H+ ions as:

Mn3+(aq) → MnO2(s) + 4H+(aq) + e(aq)

The O atoms and H+ ions are balanced by adding 2 H2O molecules as:

Mn3+(aq) + 2H2O(l) → +MnO2(s) + 4H+(aq) + e(aq) …..(i)

The reduction half equation is:

Mn3+(aq) + e → Mn2+(aq)

The oxidation number is balanced by adding one electron as:

Mn3+(aq) + e → Mn2+(aq) ….(ii)

The balanced chemical equation can be obtained by adding equation (i) and (ii) as:

2Mn3+(aq) + 2H2O(l) → MnO2(s) + 2Mn2+(aq) + 4H+(aq)

22.   (a) F exhibits only negative oxidation state of -1.

(b) Cs exhibits positive oxidation state of +1.

(c) I exhibits both positive and negative oxidation stat It exhibits oxidation states of – 1, + 1, + 3, + 5, and + 7.

(d) The oxidation state of Ne is zero. It exhibits neither negative nor positive oxidation states.

23.   The given redox reaction can be represented as:

Cl2(s) + SO2(aq) + H2O(l) → Cl(aq) + SO2-4(aq)

The oxidation half reaction is:

The oxidation number is balanced by adding two electrons as:

SO2(aq) → SO2-4(aq) + 2e

The charge is balanced by adding ions as:

SO2(aq) → SO2-4(aq) + 4H(aq) + 2e

The O atoms and H+ ions are balanced by adding 2molecules as:

SO2(aq) + 2H2O(l) → SO2-4(aq) + 4H(aq)0 + 2e ….(i)

The reduction half reaction is:

Cl2(s) → Cl(aq)

The chlorine atoms are balanced as:

The oxidation number is balanced by adding electrons

Cl2(s) + 2e → 2Cl(aq) …… (ii)

The balanced chemical equation can be obtained by adding equation (i) and (ii) as:

Cl2(s) + SO2(aq) + 2H2O(l) → 2Cl(aq) + SO2-4(aq) + 4H+(aq)

24.   In disproportionation reactions, one of the reacting substances always contains an element that can exist in at least three oxidation states.

(a) P, Cl, and S can show disproportionation reactions as these elements can exist in three or more oxidation states.

(b) Mn, Cu, and Ga can show disproportionation reactions as these elements can exist in three or more oxidation states.

25.   The balanced chemical equation for the given reaction is given as:

Thus, 68 g of NH3 reacts with 160 g of O2.

Therefore, 10g of NH3 reacts with  g of O2, or 23.53 g of O2.

But the available amount of O2 is 20 g.

Therefore, O2 is the limiting reagent (we have considered the fixed amount of Oto calculate the weight of nitric oxide obtained in the reaction).

Now, 160 g of O2  gives 120g of NO.

Therefore, 20 g of O2 gives  g of N, or 15 g of NO.

Hence, a maximum of 15 g of nitric oxide can be obtained.

26.   (a) The possible reaction between Fe3+(aq) + I(aq) is given by,

E° for the overall reaction is positive. Thus, the reaction between Fe3+(aq) and I(aq) is feasible.

(b) The possible reaction between Ag+(aq) + Cu(s) is given by,

E° positive for the overall reaction is positive. Hence, the reaction between Ag+(aq) and Cu(s) is feasible.

(c) The possible reaction between Fe3+(aq) and Cu(s) is given by,

E° positive for the overall reaction is positive. Hence, the reaction between Fe3+(aq) and Cu(s) is feasible.

(d) The possible reaction between Ag(s) and Fe3+(aq) is given by,

Here, E° for the overall reaction is negative. Hence, the reaction between Ag(s) and Fe3+(aq) is not feasible.

(e) The possible reaction between Br2(aq) and Fe2+(aq) is given by,

Here, E° for the overall reaction is positive. Hence, the reaction between Br2(aq) and Fe2+(aq) is feasible.

27.   (i) AgNO3 ionizes in aqueous solutions to form Ag+ and NO3 ions.

On electrolysis, either Ag+ ions or  molecules can be reduced at the cathode. But the reduction potential of Ag+ ions is higher than that of H2O.

Ag+(aq) + e → Ag(s); E0 = ++0.83V

2H2O(l) + 2e → H2(g) + 2OH(aq); E0 = -0.83V

Hence, Ag+ ions are reduced at the cathode. Similarly, Ag metal or H2O molecules can be oxidized at the anode. But the oxidation potential of Ag is higher than that of H2O molecules.

Ag(s) + e → Ag+(aq); E0 = -0.80V

2H2O(l) → O2(g) + 4H+(aq)+4e ; E0 = -1.23V

Therefore, Ag metal gets oxidized at the anode.

(ii) Pt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate O2. At the cathode, Ag+ ions are reduced and get deposited.

(iii) H2SO4 ionizes in aqueous solutions to give H+ and SO42- ions.

H2SO4(aq) → 2H(aq) + SO2-4(aq)

On electrolysis, either of Hions or H2O molecules can get reduced at the cathode. But the reduction potential of H+ ions is higher than that of H2O molecules.

2H+(aq) + 2e– → + H2(g)+  ; E0 = 0.0V

2H2O(aq) + 2e– → + H2(g) + 2OH(aq); E0 = -0.83V

Hence, at the cathode, H+ ions are reduced to liberate H2 gas.

On the other hand, at the anode, either of SO2-4 ions or H2O molecules can get oxidized. But the oxidation of SO2-4 involves breaking of more bonds than that of H2O molecules. Hence, SO2-4 ions have a lower oxidation potential than H2O. Thus, H2O is oxidized at the anode to liberate O2 molecules.

(iv) In aqueous solutions, CuCl2 ionizes to give Cu2+ and Cl ions as:

CuCl2(aq) → Cu2+(aq) + Cl(aq)

On electrolysis, either of Cu2+ ions or molecules can get reduced at the cathode. But the reduction potential of Cu2+ is more than that of H2O molecules.

Cu2+(aq) + 2e → Cu(aq); E0 = +0.34V

H2O(l) + 2e → H2(g) + 2OH; E0 = -0.83V

Hence, Cu2+ ions are reduced at the cathode and get deposited.

Similarly, at the anode, either of Cuor H2O is oxidized. The oxidation potential of is higher than that of Cl.

2Cl(aq) → Cl2(aq) + 2e; E0 = -1.36V

2H2O(l) → O2(aq) + 4H+(aq) + 4e; E0 = -1.23V

But oxidation of H2O molecules occurs at a lower electrode potential than that of Cl ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl ions are oxidized at the anode to liberate Cl2 gas.

28.   A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt.

The order of the increasing reducing power of the given metals is Cu < Fe < Zn < Al < Mg.

Hence, we can say that Mg can displace Al from its salt solution, but Al cannot displace Mg.

Thus, the order in which the given metals displace each other from the solution of their salts is given below:

Mg>Al> Zn> Fe,>Cu.

29.   The lower the electrode potential, the stronger is the reducing agent. Therefore, the increasing order of the reducing power of the given metals is Ag < Hg < Cr < Mg < K.

30.   The galvanic cell corresponding to the given redox reaction can be represented as:

Zn | Zn2+(aq) || Ag(a) | Ag

(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn2+ and the leaving electrons accumulate on this elect

(ii) Ions are the carriers of current in the

(iii) The reaction taking place at Zn electrode can be represented as:

Zn(s) → Zn2+(aq) + 2e

And the reaction taking place at Ag electrode can be represented as:

Ag+(aq) + e → Ag(s).

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